Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-6x+y &= -4 \\ 8x+3y &= 1\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $8x = -3y+1$ Divide both sides by $8$ to isolate $x$ $x = {-\dfrac{3}{8}y + \dfrac{1}{8}}$ Substitute this expression for $x$ in the first equation. $-6({-\dfrac{3}{8}y + \dfrac{1}{8}}) + y = -4$ $\dfrac{9}{4}y - \dfrac{3}{4} + y = -4$ Simplify by combining terms, then solve for $y$ $\dfrac{13}{4}y - \dfrac{3}{4} = -4$ $\dfrac{13}{4}y = -\dfrac{13}{4}$ $y = -1$ Substitute $-1$ for $y$ in the top equation. $-6x- 1 = -4$ $-6x-1 = -4$ $-6x = -3$ $x = \dfrac{1}{2}$ The solution is $\enspace x = \dfrac{1}{2}, \enspace y = -1$.